CMR: câu a) 2018a-2019b / 2019c+2020d = 2018c - 2019d / 2019a+2020b
câu b) a^2 + c^2 / b^2 + d^2 = a/bd
cho \(\dfrac{a}{b}=\dfrac{c}{d}\)Chứng minh rằng
\(\dfrac{2018a-2019b}{2019c+2020d}\)=\(\dfrac{2018c-2018c}{2019a+2020b}\)
Sửa đề: \(\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2020a}{2020b}=\dfrac{2020c}{2020d}=\dfrac{2019a}{2019c}=\dfrac{2019b}{2019d}=\dfrac{2018a}{2018c}=\dfrac{2018b}{2018d}=\dfrac{2018a-2019b}{2018c-2019d}=\dfrac{2019a+2020b}{2019c+2020d}\\ \Leftrightarrow\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)
\(\dfrac{2018a-2019b}{2019c-2020d}=\dfrac{2018c-2018c}{2019a+2020b}\)
Sao .... ;-; ;-;
\(Cho:\frac{a}{2b}+\frac{b}{2c}+\frac{c}{2d}+\frac{d}{2a}\)\(\left(a,b,c,d>0\right)\)Tính:\(\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{c+b}\)
\(Cho\frac{a}{b}=\frac{c}{d}.CMR:\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019c^2+2020d^2}{2019c^2-2019d^2}\)
#Ttql
#Duongg
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\Rightarrow\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019b^2k^2+2020b^2}{2019b^2k^2-2020b^2}\)
\(=\frac{2019k^2+2020}{2019k^2-2020}\)(1)
và\(\Rightarrow\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019d^2k^2+2020d^2}{2019d^2k^2-2020d^2}\)
\(=\frac{2019k^2+2020}{2019k^2-2020}\)(2)
Từ (1) và (2) suy ra \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}\)\(=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\left(đpcm\right)\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng ta có tỉ lệ thức sau :
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Dăm ba mấy bài đặt k:v
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Ta có:
\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018b^2k^2+2019b^2}{2018b^2k^2-2019b^2}=\frac{b^2\left(2018k^2+2019\right)}{b^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)
\(\frac{2018c^2+2019d^2}{2018c^2-2019d^2}=\frac{2018d^2k^2+2019d^2}{2018d^2k^2-2019d^2}=\frac{d^2\left(2018k^2+2019\right)}{d^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)
Từ đó \(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}.Chứngminh\frac{2018a-2019b}{2018c+2019d}=\frac{2018c-2019d}{2018a+2019b}\)
Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{2018a}{2018c}=\frac{2019b}{2019d}\)
Áp dụng t/c DTSBN : \(\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2018a-2019b}{2018c-2019d}=\frac{2018a+2019b}{2018c+2019d}\)
Cái này đến đây là đề sai nhé ! Đề phải cho là C/m cái (2018a-2019b).(2018c+2019d) = (2018a-2019b)(2018c+2019d) mới đúng
1.Tính:
\(\left(\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\right)\times\frac{1-3-5-7-...-49}{89}\)
2.Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\). Tính: \(A=\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{b+c}\)
3.Tìm x biết:\(\left(x-1\right)\left(x-3\right)< 0\)
2.
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)
\(\Rightarrow a=\frac{2b}{2}=b;b=\frac{2c}{2}=c;c=\frac{2d}{2}=d;d=\frac{2a}{2}=a\)
\(\Rightarrow a=b=c=d\)
Ta có : \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)
\(=\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}\)
\(=\frac{4a}{2a}=2\)
3.
\(\left(x-1\right)\left(x-3\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-1< 0\\x-3>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 1\\x>3\end{cases}}\)( loại ) hoặc \(\hept{\begin{cases}x>1\\x< 3\end{cases}}\)
Vậy \(1< x< 3\)
Đặt \(A=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\)
Ta có : \(5\times A=\frac{5}{4\times9}+\frac{5}{9\times14}+\frac{5}{14\times19}+...+\frac{5}{44\times49}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}=\frac{1}{4}-\frac{1}{49}\)
\(=\frac{49}{196}-\frac{4}{196}=\frac{45}{196}\)
\(\Rightarrow A=\frac{9}{196}\)
Đặt \(B=1-3-5-7-...-49=1-\left(3+5+...+49\right)\)
Đặt \(C=3+5+...+49\) ( khoảng cách là 2 )
Số số hạng là : \(\left(49-3\right):2+1=24\)
Tổng C là : \(\left(49+3\right)\times24:2=624\)
\(\Rightarrow B=1-264=-623\)
Vậy \(A=\frac{9}{196}\times\frac{-623}{89}=\frac{-9}{28}\)
Dòng cuối cùng mình không chắc là đúng nhé !
\(\left(x-1\right)\left(x-3\right)< 0\)
=> x-1 và x-3 trái dấu
mà x-1>x-3 nên ta có:
\(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow}-1< x< 3}\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
vậy x \(\in\left\{-2;-1;0;1;2\right\}\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR \(\frac{2017-2018b}{2018a+2019b}=\frac{2017c-2018d}{2018c+2019d}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2017a-2018b}{2017c-2018d}=\frac{2018a+2019b}{2018c+2019d}\)
<=>\(\left(2017a-2018b\right)\left(2018c+2019d\right)=\left(2018a+2019b\right)\left(2017c-2018d\right)\)
<=>\(\frac{2017a-2018b}{2018a+2019b}=\frac{2017c-2017d}{2018x+2019d}\)(đpcm)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng ta co tỉ thức sau :
\(\frac{2018a^{2\:}+2019b^2}{2018b^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)
Đặt bằng k nhé các bạn , giúp mình nhanh lên ạ
Nhanh lên ạ
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{2018a^2}{2018c^2}=\frac{2019b^2}{2019d^2}=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}=\frac{2018a^2-2019b^2}{2018c^2-2019d^2}\)
\(\Rightarrow\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\left(dpcm\right)\)
Cho a,b,c,d thỏa mãn $\frac{a}{b}$ =$\frac{b}{c}$ =$\frac{c}{d}$ =$\frac{d}{a}$
CMR:($\frac{2019b+2020c-2021d}{2019c+2020d-2021e}$)^3=$\frac{a^2}{bc}$
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)